saharon shelah- characterizing an ℵ-epsilon saturated model of superstable ndop theories by its...

8/3/2019 Saharon Shelah- Characterizing an -epsilon Saturated Model of Superstable NDOP Theories by its L,-epsilon Th

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CHARACTERIZING AN -SATURATEDMODEL OF SUPERSTABLE NDOP

THEORIES BY ITS L, -THEORY

SH401

Saharon Shelah

The Hebrew University of JerusalemEinstein Institute of Mathematics

Edmond J. Safra Campus, Givat RamJerusalem 91904, Israel

Department of MathematicsHill Center-Busch Campus

Rutgers, The State University of New Jersey110 Frelinghuysen Road

Piscataway, NJ 08854-8019 USA

MSRIBerkeley, CA USA

Abstract. Assume a complete countable first order theory is superstable withNDOP. We knew that any -saturated model of the theory is -prime over anon-forking tree of small models and its isomorphism type can be characterized byit L,(dimension quantifiers)-theory; or if you prefer - appropriate cardinal invari-ants. We go here one step further providing cardinal invariants which are as finitaryas seems reasonable.

Partially supported by the United States-Israel Binational Science Foundation and the NSF, andI thank Alice Leonhardt for the beautiful typing.Revised with proofreading for the JournalDone Oct/89Publ No. 401Latest Revision - 05/Dec/20

Typeset by AMS-TEX

1


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2 SAHARON SHELAH

Introduction

After the main gap theorem was proved (see [Sh:c]), in discussion, Harrington

expressed a desire for a finer structure - of finitary character (when we have a struc-ture theorem at all). I point out that the logic L,0(d.q.) (where d.q. stands fordimension quantifier) does not suffice: suppose; e.g. for T = T h ( 2, En)nord is nonempty closed under initial segments

(b) M M is -saturated

(c) I M M

(d) if = I then M is -prime over M {a} and tp(a, M) is

orthogonal to M for and (the last is not essential but clarifies)(e) M : I is non-forking enough: for every I the set {a :

SucI()} M is independent over M .

The point is that if = , M , a are chosen then to a large extent M, a : I is determined. But the amount of to a large extent which suffices in[Sh:c], is not sufficient here, we need to find a finer understanding. In particular,we certainly do not like to know (M, a). So we consider a pair (A, B) whereA M , A {a} B M, stp(B, A) stp(B, M) and we try to define thetype of such pairs in a way satisfying:

(a) it can be impressed in our logic L,(b) it expresses the essential information in M, a : I.

To carry out the isomorphism proof we need: (1.27) the type of the sum is the sumof types (infinitary types) assuming first order independence. The main point of theproof is to construct an isomorphism between M1 and M2 when M1 L, (d.q.)M2, T h(M) = T where T and L, (q.d.) are as above. So by [Sh:c, X] it is


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CHARACTERIZING MODELS OF NDOP THEORIES 3

enough to construct isomorphic decompositions. The construction of isomorphicdecompositions is by approximations, in stage n, n levels of the decompositiontree are approximated, i.e. we have In

nOrd and an, M for In, = 1, 2

such that tp(an,1

0an,1

1 . . . an,1 , , M) = tp(an,2

0an,2

1 . . . an,2 , , M2) with a,being -finite, so in stage n + 1, choosing an+1, we cannot take care of all types

an+1, a, so addition theorem takes care. So though we are thinking on -

decompositions (i.e., the Ms are -saturated), we get just a decomposition.In the end of 1 (in 1.37) we point up that the addition theorem holds in fuller

generalization. In the second section we deal with a finer type needed for shallowT, in the appendix we discuss how absolute is the isomorphism type.

Of course, we may consider replacing -saturated models of an NDOP su-perstable countable T by models of an NDOP 0-stable countable T. But theuse of-finite sets seems considerably less justifiable in this context, it seems morereasonable to use finite sets, i.e., L,0(d.q.). But subsequently Hrushovski andBouscaren proved that even if T is 0-stable, L,0(d.q.) is not sufficient to char-acterize models ofT up to isomorphism. This is not sufficient even if one considersthe class of all -saturated models rather than all models. The first example is0-stable shallow of depth 3, and the second one is superstable (non 0-stable),NOTOP, non-multi-dimensional.

If we deal with -saturated models of shallow (superstable NDOP) theories T,we can bound the depth of the quantification = DP(T); i.e. L, suffice.

We assume the reader has a reasonable knowledge of [Sh:c, V,1,2] and mainly[Sh:c, V,3] and of [Sh:c, X].

Here is a slightly more detailed guide to the paper. In 1.1 we define the logicL, and in 1.3 give a back and forth characterization of equivalence in this logicwhich is the operative definition for this paper.

The major tools are defined in 1.7, 1.11. In particular, the notion of tp definedin 1.5 is a kind of a depth look-ahead type which is actually used in the finalconstruction. In 1.28 we point out that equivalence in the logic L, impliesequivalence with respect to tp for all . Proposition 1.14 contains a number ofimportant concrete assertions which are established by means of Facts 1.16-1.23.In general these explain the properties of decompositions over a pair

BA

. Claim

1.27 (which follows from 1.26) is a key step in the final induction. Definition 1.30

establishes the framework for the proof that two -saturated structures that havethe same tp are isomorphic. The induction step is carried out in 1.35.

I thank Baldwin for reading the typescript pointing needed corrections and writ-ing down some explanations.

-1.1 Notation: The notation is of [Sh:c], with the following additions (or reminders).If = then we let = ; for I a set of sequences ordinals we let


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4 SAHARON SHELAH

SucT() = { : for some , = I}.We work in Ceq and for simplicity every first order formula is equivalent to a relation.

(1) means orthogonal (so q is p means q is orthogonal to p),remember p A means p orthogonal to A; i.e. p q for everyq S(ac(A)) (in Ceq)

(2) a means almost orthogonal

(3) w means weakly orthogonal

(4) aB

and a/B means tp(a, B)

(5) AB

or A/B means tp(A, B)

(6) A + B means A B

(7)

A{Bi : i < } means {Bi : i < } is independent over A

(8) AB

C means {A, C} is independent over B

(9) {Ci : i < } is independent over (B, A) means that1

j < tp(Cj , {Ci B : i = j}) does not fork over A

(10) regular type means stationary regular type p S(A) for some A

(11) for p S(A) regular and C a set of elements realizing p, dim(C, p) isMax{|I| : I C is independent over A}

(12) ac(A) = {c : tp(c, A) is algebraic}

(13) dc(A) = {c : tp(c, A) is realized by one and only one element}

(14) Dp(p) is depth (of a stationary type, see [Sh:c, X,Definition 4.3,p.528,Definition4.4,p.529]

(15) Cb(p) is the canonical base of a stationary type p (see [Sh:c, III,6.10,p.134])

(16) B is -atomic over A if for every finite sequence b from A, for some findA0 A we have stp(b, A0)) stp(b, A), equivalently for some -finiteA0 acl(A) we have tp(b, A0)) tp(b, acl(A)).

1Actually by the non-forking calculus this is equivalent to {Ci : i } is independent over Awhere we let C = B.


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1 -saturated models

We first define our logic, but as said in 0, we shall only use the condition from1.4. T is always superstable complete first order theory.

1.1 Definition. 1) The logic L, is slightly stronger than L,0 , it consists ofthe set of formulas in L,|T|+ such that any subformula of of the form (x) isactually the form

(x0, x1)

1(x1, y) &

i


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6 SAHARON SHELAH

a A instead of a >(A).2) A is -finite, if for some a >A, A = ac(a). (So for stable theories a subsetof an -finite set is not necessarily -finite but as T is superstable, a subset of an

-finite set is -finite as ifB ac(a), b B is such that tp(a, B) does not fork overb, then trivially ac(b) A and if ac(b) = B, tp(B, ab) forks over B, hence([Sh:c, III,0.1]) tp(a, B) forks over b, a contradiction.So if ac(A) = ac(B), then A is -finite iff B is -finite).3) When T is superstable by [Sh:c, IV,Table 1,p.169] for F = Fa0 , all the axiomsthere hold and we write instead of F and may use implicitly the consequencesin [Sh:c, IV,3].

We may instead Definition 1.1 use directly the standard characterization from 1.4;as actually less is used we state the condition we shall actually use:

1.4 Claim. For models M1, M2 of T we have M1 L, (d.q.) M2 iffthere is a non-empty familyF such that:

(a) each f F is an (M1, M2)-elementary mapping, (so Dom(f) M1,Rang(f) M2)

(b) forf F, Dom(f) is -finite (see 1.3(2)) above)

(c) if f F, a M ( = 1, 2) then for some g F we have:f g and ac(a1) Dom(f) and ac(a2) Rang(f)

(d) if f {a1, a2} F and tp(a1, Dom(f)) is stationary and regularthen dim({a1

1 M

1: f {a1

1, a

2} F}, M

1)

= dim({a12 M2 : f {a1, a12} F}, M2).

Our main theorem is

1.5 Theorem. Suppose T is countable (superstable complete first order theory)with NDOP.Then

(1) the L,(d.q.) theory of an -saturated model characterizes it up to iso-morphism.

(2) Moreover, if M1, M2 are -saturated models of T (so M Ceq) and

M0,M1of 1.4 holds, then M1, M2 are isomorphic.

By 1.4, it suffices to prove part (2).


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The proof is broken into a series of claims (some of them do not use NDOP, almostall do not use countability; but we assume T is superstable complete all the time(1.7(1)).

1.6 Discussion: Let us motivate the notation and Definition below.Recall from the introduction that we are thinking of a triple (M , N , a) which

may appear in -decomposition M, a : I of N, in the sense that for some I\{} we have (M, M, a) = (M, M, a) so M, M

are -saturated, a M\M, M is -prime over M + a and tp(a, M) is regular. But this is too largefor us hence we consider an approximation (A, B) where A M(= M), A B M(= M)), a = a B and B/M(= B/M) does not fork over A. Wewould like to define the -type of (A, B) in N, which tries to say something on thedecomposition above (M, M, a) = (M, M, a), i.e., on M, a : I. Thereare two natural successor of (A, B) we may choose in this context: the first 1.7below replaces (A, B) to (A, B) such that A A M(= M), B B

M(=M) and (as M

is -prime over M + a) we have stp(B, A B) stp(B, M),

so tp(B, A B) is almost orthogonal to A; we can think of this as advancing inthe same model; in other words as A, B are -finite, we have to increase them inorder to capture even (M, M). This is formalized by a in Definition 1.7 below.

The second is to pass from (M, M, a) to (M, M, a) for some an immediatesuccessor (in I) of I. So the old B is included in the new A and B = A {a}where tp(a, A) is regular and is orthogonal to A (as in the decomposition we requiretp(a, M)(M when

). This is formalized by b in Definition 1.7 below.

1.7 Definition. 1) = {(A, B) : A B are -finite}. Let(M) = {(A, B) : A B M}.2) For members (A, B) of we may also write

BA

; if A B we mean

BAA

.

3)B1A1

aB2A2

(usually we omit a) if (both are in and)

A1 A2, B1 B2, B1A1

A2 andB2

B1+A2a A2.

4)B1A1

bB2A2

if A2 = B1, B2\A2 = b and

bA2

is regular orthogonal to A1.

5) is the transitive closure of a b. (So it is a partial order, whereas ingeneral a b and b are not).6) We can replace A, B by sequences listing them (we do not always strictly distin-

guish).

Remark. The following observation may clarify.

1.8 Observation. IfB1A1

B2A2

then we can find B : n and


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8 SAHARON SHELAH

c : 1 < n for some n 1, satisfyingB1A1

bB1B0

, c B

+1,

cB

regular,

B+1c+B

a B, A2 = B

n1, B2 = B

n.

Remark. 1) Note that actually a is transitive. This means that in a sense b isenough, a inessential.2) We may in 1.7(4) use b = c, does not matter.

Proof. By the definition of there are k < andB

A

for k such that:

B

A

x()

B+1

A+1

for k and x() {a,b} and

B0

A0

=B1A1

,Bk

Ak

=B2A2

and

without loss of generality x(2) = a,x(2 + 1) = b. Let N0 C be -prime

over such that A0 N0, B0 A0

N0 and f0 = idA0 . We choose by induction on

k, N+1, f+1 such that:

(a) Dom(f+1) = B

(b) N N+1

(c) if x() = b then f+1 is an extension of f (which necessarily has domainA, check) with domain B

such that f(B)

f(A)

N and N+1 is -prime

over N f(B)

(d) if x() = a, then f+1 maps A into N1, B

into N and N+1 = N.

This is straightforward. Now on N : k + 1 we repeat the argument (ofchoosing B : n) in the proof of 1.14(6) above, i.e., choose B

N bydownward induction on large enough as required. 1.8

1.9 Definition. 1) We define tp[BA

, M] (for A B M, A and B are

-finite and is an ordinal) and S(BA

, M),S(A, M) andS

r(BA

, M),Sr(A, M)

by induction on (we mean simultaneously; of course, we use appropriate vari-

ables):

(a) tp0[BA

, M] is the first order type of A B

(b) tp+1[BA

, M] = the triple Y1,A,B,M, Y

2,A,B,M, tp(

BA

, M)

where:

Y1,A,B,M =:

tp[B

A

, M] : for some A, B we have

BA

aB

A

(M)

,


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and Y2,A,B,M =:

, M,B : S

r(B, M)

where M,B = dim d : tp[B+dB , M] = , B:(c) for a limit ordinal, tp[

BA

, M] = tp[

BA

, M] : <

(this includes = , really M+ suffice).

(d) S(A, M) =

tp[BA

, M] : for some B such that B M,

andBA

(M)

(e) Sr(BA

, M) =

tp[B+cB

, M] : for some c M we have

cB A and cB is regular(f) Sr(A, M) =

tp[A+cA

, M] : c M and cA regular

.

2) We define also tp[A, M], for A an -finite subset of M:

(a) tp0[A, M] = first order type of A

(b) tp+1[A, M] is the triple Y1,A,M, Y

2,A,M, tp[A, M] where

Y1,A,M =: S(A; M) and

Y2,A,M =:, dim{d M : tp[A+dA , M] = } : Sr(A, M)

(c) tp[A, M] = tp(A, M) : <

3) tp[M] = tp[, M].

1.10 Discussion: Clearly tp[BA

, M] is intended, on the one hand, to be expressible

by our logic and, on the other hand, to express the isomorphism type of M in thedirection of

BA

. To really say it we need to go back to the -decompositions of

M, a central notion of [Sh:c, Ch.X].

For the readers benefit, by the referee request, let us review informally theproof in [Sh:c, Ch.X]. Let M be an -saturated model, and we choose M : I nOrd, a : I n+1 by induction on n. For n = 0, of course,I 0Ord = {}, we let N M be -prime over and let I a maximalsubset of{c M : tp(c, N) regular} which is independent over N, let a : < |I| list I. Similarly for n + 1, I n

+1Ord let N M be -prime over M + a, let I be a maximal subset of {c M : tp(c, M) is regular


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10 SAHARON SHELAH

orthogonal to M} independent over N. Lastly, let c : < |I| list Iand let I n+1Ord = { < >: I nOrd and < |I|}.

To carry this we use the existence of -prime models (and the local character

of indpendent). Also looking at the set {M : I}, its first order type isdetermined by the non-forking calculus. In fact, for any I\{}, the set{N : I}, {N : ( ) and I} are independent over N. Let N Mbe -prime over {N : I}, now ifM = N we are done decomposing M, if notsome c M\N realize a regular type (we use density of regular types). By NDOP,the tp(c, N) is not orthogonal to some N. Choose of minimal length hence tp(c, M) N . By properties of regular types, without loss of generality tp(c, N)

does not fork over N, so we get a contradiction to the maximality of {a : SucI()} (this explains the role ofP in Definition 1.11(5) below).

We are interested in the possible trees N : I.

Now the tree determines M up to isomorphism, but there are incidentalchoices, so two trees may give isomorphic models (for investigating the numberof non isomorphic models it is enough to find sufficiently pairwise far trees I).

Here we like to get exact information and in as finitary way as we can. So wereplace (M, M, a) by

BA

, where A M, A + a B M, tp(B, M)

does not fork over A.Now for I\{} we are interested in the possible trees N : I, over

(N , N, a). But not only different trees may be equivalent (giving isomorphic

-prime models) but the other part of the tree, N : I but ( ) mayapriori cause non equivalent trees to contribute the same toward understanding M.This is done in [Sh:c, Ch.XII], but here we have to deal with -finite A, B.

The following claim 1.11 really does not add to [Sh:c, Ch.X], it just collects therelevant information which is proved there, or which follows immediately (paricu-larly using the parameter (A, B)). We allow here a/M - to be not regular, butthis is not serious: we can here deal exclusively with this case and we can omitthis requirement in [Sh:c, Ch.X]; however, this does not eliminate the use of regulartypes (in the proof that M is -prime over every -decomposition of it).

1.11 Definition. 1) N, a : I is an -decomposition inside M above (or

for over) the pairBA

(but we may omit the ) if:

(a) I a set of finite sequences of ordinals closed under initial segments

(b) , 0 I, I\{} 0 , let I = I\{}, really a is meaningless

(c) A N, B N0, NA

B and dc(a0) dc(B),

(d) if = I then N is -primary over N a , N is -primeover A


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(e) for I such that k = g() > 1 the type a/N(k1) is orthogonal toN(k2)

(f) N N

(g) M is -saturated and N M for I

(h) if I\{}, then {a : SucI()} is (a set of elements realizing overN types orthogonal to N and is) an independent set over N.

2) We replace inside M by of M if in addition

(i) in clause (h) the set is maximal.

3) N, a : I is an -decomposition inside M if (a), (d), (e), (f), (g), (h) ofpart (1) holds and in clause (h) we allow = (call this (h)+). We add over Aif A M.4) N, a : I is an -decomposition of M if in addition to 1.11(3) and wehave the stronger version of clause (i) of 1.11(2) by including = , i.e. we have:

(i)+ for I, the set {a : SucI()} is a maximal subset of M independentover N .

We may add over A if A M.5) IfN, a : I is an -decomposition inside M we let

P(N, a : I, M) =

p S(M) : p regular and for some I\{} we have

p is orthogonal to N but not to N.As said earlier it is natural to use regular types.

1.12 Definition. 1) We say that N, a : I, an -decomposition inside M,is J-regular if J I and:

() for each I\J there2 is c such that a ac(N + c)

cN

is regular and if = thena

N+ca N().

2) We say N, a : I is a regular -decomposition inside M [of M] if it isan -decomposition inside M [of M] which is -regular.3) We say N, a : I is a regular -decomposition inside M [of M] overBA

if it is an -decomposition inside M [of M] over

BA

which is {}-regular.

2without loss of generality c = a


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12 SAHARON SHELAH

1.13 Claim. 1) Every -saturated model has an -decomposition (i.e. of it).2) If M is -saturated, N, a : I is an -decomposition inside M, thenfor some J, and N, a for J\I we have: I J and N, a : J is an

-decomposition of M (even a (J\I)-regular one).3) If M is -saturated, N, a : I is an -decomposition of M then M is

-prime and -minimal3 over

I

N; if in addition N, a : {, 0} is an

-decomposition inside M aboveBA

, then N, a : I & ( = 0 )

is an - decomposition of M aboveBA

.

4) If N, a : I is an -decomposition inside M aboveBA

, then it is an

-decomposition inside M.5) If N, a : I is an -decomposition inside M [above

BA

], I,

[ I\{}], = Min{ : / I}, =: , a M\N,aN

is orthogonal

to M if if = , N M is -primary over N + a and a

N

I

N

(enough to demand {a : = and I} is independent over a/N) then

N, a : I {} is an -decomposition inside M [overBA

].

6) Assume N, a : I is an-decomposition ofM, if p is regular (stationary)and is not orthogonal to M (e.g. p S(M)) then for one and only one I, thereis a regular (stationary) q S(N) not orthogonal to p such that: if

is welldefined (i.e. = ), then p N.

7) Assume I =


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10) Assume that for = 1, 2 that N, a : I is an-decomposition inside M,

and for I the functionf is an isomorphism fromN1 onto N

2 and f

f. ThenIf is an elementary mapping; if in addition N, a

: I is an -

decomposition of M (for = 1, 2) thenI

f can be extended to an isomorphism

from M1 onto M2.11) If N, a : I is an- decomposition inside M (above

BA

) and M M

is -prime overI

N then N, a : I is an -decomposition of M

(aboveBA

).

12) If N, a; I is an -decomposition inside M/ of M [aboveMA

] and

a N and N is -prime over N + a for I\{} [and a

= a

or at least dcl(a) dcl(B) ] thenN, a : I is an -decomposition inside

M/ of M [aboveBA

].

Proof. 1), 2), 3), 5), 6), 9)-12). Repeat the proofs of [Sh:c, X]. (Note that herea/N is not necessarily regular, a minor change).4), 7) Check.8) As Dp(p) > 0 p is trivial, by [Sh:c, ChX,7.2,p.551] and [Sh:c, ChX,7.3]. 1.13

We shall prove:

1.14 Claim. 1) If M is -saturated,BA

(M), then there is N, a : I,

an -decomposition of M aboveBA

.

2) Moreover if N, a : I satisfies clauses (a) (h) of Definition 1.11(1), wecan extend it to satisfy clause (i) of 1.11(2), too.

3) IfN, a : I is an-decomposition of M aboveBA

, M M is -prime

overI

N then:

(a) N : I is a -decomposition of M

(b) we can find an -decomposition N, a : J of M such that J I and[ J\I ( = and 0 )], moreover the last phrase follows fromthe previous ones.

4) If in 3)(b) the set J\I is countable (finite is enough for our applications), thennecessarily M, M are isomorphic, even adding all members of an -finite subsetof M as individual constants.5) If N, a : I is an -decomposition of M above

BA

, I J and


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14 SAHARON SHELAH

N, a : J is an -decomposition of M, M M is -prime over

I

N

and BA

B1A1 and B1 M and c M and

cB1

A1 andcB1

is (stationary and)

regular then

() cB1

{N:J\I}N() cB1 is not orthogonal to some p P(N, a : I, M

).

6) If N, a : I is an -decomposition of M aboveBA

, then the setP =

P(N : I, M) depends onBA

and M only (and not on N : I or M

when M M is -prime over {N : I}), recalling:

P = P(N : I, M) = p S(M) :p regular and for some I\{}, we have :

p is orthogonal to N but not to N

.

So letP(BA

, M) =: P(N : I, M).

(7) If BA is regular of depth zero or justbA a

BA ,

bA regular of depth zero and M is

-saturated and B M then

(a) for any , we have tp(BA, M) depend just on tp0(BA, M)

(b) ifBA

B

A

(M) then tp(

B

A

, M) depends just ontp0(

BA

, M) (and

(A,B,A, B) but not on M).

8) For < , from tp(BA

, M) we can compute tp(

BA

, M).

9) If f is an isomorphism from M1 onto M2, A1 B1 are -finite subsets of M1and f(A1) = A2, f(B1) = B2 then

tp(

B1A1

, M1) = tp(

B2A2

, M2)

(more pedantically tp(B2A2

, M2) = f[tp(

B1A1

, M1)] or considered the A, B as

indexed sets).

We delay the proof (parts (1), (2), (3) are proved after 1.22, part (4), (6) after 1.23,and after it parts (5), (7), (8). Part (9) is obvious.


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1.15 Definition. 1) IfBA

(M), M is -saturated let PM(BA)

be the set P from

Claim 1.14(6) above (by 1.14(6) this is well defined as we shall prove below).

2) LetP

(BA) = p : p is (stationary regular and) parallel to some p PCeq

(BA).1.16 Definition. If N, a : J is a decomposition inside C for = 1, 2 we

say that N1 , a : J direct N

2 , a : J if:

(a) N1 N2

(b) N2

N1

{a : J}

(c) for J\{}, N2 is -prime over N1 N

2.

1.17 Claim. 1) M is -prime over A iff M is -primary over A iff M is -saturated, A M, M is -atomic over A (see -1.1(16)) for every I M in-discernible over A we have: dim(I, M) 0 iff M is -saturated, A M, M is-atomic overA and for every finite B M and regular (stationary) p S(AB),we have dim(p,M) 0.2) If N1, N2 are -prime over A, then they are isomorphic over A.

Proof. By [Sh:c, IV,4.18] (see Definition [Sh:c, IV,4.16], noting that we replace Fa0by and that part (4) there disappears when we are speaking on Fa0). 1.16

However, we need more specific information saying that minor changes preservebeing -prime; this is done in 1.18 below, parts of it are essentially done in [Sh225] but we give full proof.

1.18 Fact. 0) If A is countable, N is -primary over A then N is -primaryover .

1) If N is -prime over , A countable, N+ is -prime over N A then N+ is-prime over .

2) If Nn : n < is increasing, each Nn is -prime over or just -constructible over and N is -prime over

n


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16 SAHARON SHELAH

of tp(b, a) then dim(q, C a) = dim(tp(b, a), N) = 0.2B) If C ab N and a/b is a regular type orthogonal to C and q Sg(a)(N)is a non-forking extension of a/b then dim(p (C + b), N) dim(a/b, N)

dim(p (C+ b), N) + 0; moreover, dim(p (C+ b), N) dim(a/b, N) < dim(p (C+ b), N)+ + 0.3) If N2

N0

N1, each N is -saturated, N2 is -prime over N0 a, and N3 is

-prime over N2 N1 then N3 is -prime over N1 a.4) If N1 N2 are -primary over then for some -saturated N0 N1 (neces-sarily -primary over ) we have: N1, N2 are isomorphic over N0.5) In part (4), if A N1 is -finite then we can demand A N0.6) IfM0 is -saturated, A

M0

B, M1 is -primary over M0 A then M1

M0

B.

7) Assume N0 N1 N2 are -saturated, N2 is -primary over N1 + a andaN1 N0 (and a / N1). If N0 N0, N0 N1 N1, N1

N0

N0 and N1 is -

primary over N0 N1, A

1 N

1, A

2 N2 are -finite and tp(A

2, N1) does not fork

over A1, then we can find a, N2 such that: N

2 is -saturated, -primary over

N1 + a, N1 N

2 N2, N1

N1

N2 and N2 is -primary over N1 N2 and A

2 N

2.

8) Assume N0 N0 N1 and a N1 and N1 is -prime over N0 + a andaN0

N0and A0 N

0, A

1 N1 are -finite and tp(A

1, N0) does not fork over A

0 then we

can find a, N1 such that a N, N0 N

1 N1, N

1

N0

N0, N1 is -prime over

N0 + a and N1 is -prime over N0 + N

1 and A

1 N

1.9) If N1 is -prime over and A B N1 and A, B are -finite, then we can

find N0 such that: A N0 N1, N0 is -prime over , A N0, BA

N0, and N1

is -prime over N0 B.10) IfN0 is -prime over A and B N0 is -finite, then N0 is -prime over A B(and also over A if A A acl(A)).

1.19 Remark. In the proof of 1.18(1)-(6),(10) we do not use T has NDOP.

Proof. 0) There is {a : < }, a list of members of N in which every member

of N\A appears such that for < () we have: tp(a, A {a : < }) is-isolated (which means just Fa0 -isolated).[Why? by the definition of N is -primary over A). Let {bn : n < } list A(if A = the conclusion is trivial so without loss of generality A = , hence wecan find such a sequence bn : n < ). Now define

= + and b+ = a


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for < . So {b : < } lists the elements of N (possibly with repetitions,

remember A N and check). We claim that tp(b, {b : < }) is Fa0 -isolatedfor < .

(Why? if , let = (so < ), now the statement above meanstp(a , A {a : < }) is Fa0-isolated which we know; if < this statementis trivial)]. By the definition of Fa0-primary, clearly b : < + exemplifythat N is Fa0- primary over .1) Note

()1 if N is -primary over and A N is finite then N is -primary over A[why? see [Sh:c, IV,3.12](3),p.180 (of course, using [Sh:c, IV, Table 1,p.169]for Fa0 ]

()2 if N is -primary over , A N is finite and p Sm(N) does not forkover A and p A is stationary then for some {a : < } we have: a N

realize p, {a : < } is independent over A and p (A N realize stp(a, B1a) and letBc = B

1c bc, so tp(c, N) does not fork over Bc and tp(c, Bc) is stationary, hence

we can find ac : < } as in ()2 (for tp(c, Bc)). LetA = {Bc : c >A} {ac : c

>A and < }, so A is a countable subsetof N and tp(A, A

) tp(A, N) = stp(A, N). As N is -primary over we canfind a sequence d : <

and w : < such that N = {d : <

}and w is finite and stp(d, {d : w}) stp(d, {d : < }) and < d = d.

We can find a countable set W

such that A

{d : W} and W w W. Let A = {a : W}. By [Sh:c, IV,2,3] without loss of

generality W is an initial segment of .Easily

< & / W stp(d, {d : w) stp(d, A {d : < }).


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18 SAHARON SHELAH

As N+ is -primary over NA we can find a list {d : [, )} ofN+\(NA)

such that tp(d, N A {d : [, )}) is -isolated. Sod : / W, <

exemplifies that N+ is -primary over A A, hence by

1.18(0) we know that N+

is -primary over .2) We shall use the characterization of N is Fa0 -prime over A in 1.17, moreexactly we use the last condition in 1.17(1) for A = , M = N. Clearly N is -

saturated (as it is -prime overn


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and < < B B then


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20 SAHARON SHELAH

Now use 1.17(1) to deduce: N1 is Fa

-prime over hence (by uniqueness of-primemodel, 1.17(2)) N1

= N1.By renaming without a loss of generality N1 = N1. Now

()() (N1, c)cN0 , (N2, c)cN0 are -saturated and

() if a C, b N, a/b a regular type and ab

(N0 + b) (for = 1 or = 2),

thendim(a/(b N0), N) = 0.

[Why? Remember that we work in (Ceq, c)cN0 . The -saturated follows fromthe second statement.Note: dim(a/(b N0), N) dim(a/b, N) 0 (first inequality by monotonicity,second inequality by 1.17(1) and the assumption N is -prime over ). If a/b

is not orthogonal to N0 then for some i < we have pi (a/b) so easily (usingN is -saturated) we have dim(a/(b N0), N) = dim(pi, N) Ii = 0;so together with the previous sentence we get equality. Lastly, if a/b N0 bypart (2B) of 1.18, we have dim(a/(b N0), N) < 0 dim(a/b, N) < 0 whichcontradicts the assumption N is -saturated.] So we have proved () hence by1.17(1) we get N1, N2 are isomorphic over N

0 as required.

5) This is proved similarly as if N is -prime over A and B N is -finite thenN is -prime over A + B and also over A

if A + B A acl(A + B), see part(10).6) By [Sh:c, V,3.2].7) First assume that A

2 N1 and a/N1 is regular. As N1 is -prime over N0 N

1and as T has NDOP (i.e. does not have DOP) we know (by [Sh:c, X,2.1,2.2,p.512])that N1 is -minimal over N0 N

1 and

aN1

is not orthogonal to N0 or to N1.

But a/N1 N0 by an assumption, so a/N1 is not orthogonal to N1 hence thereis a regular p S(N1) not orthogonal to

aN1

hence (by [Sh:c, V,1.12,p.236]) p

is realized say by a N2. By [Sh:c, V,3.3], we know that N2 is -prime overN1 + a

. We can find N2 which is -prime over N1 + a

and N2 which is -primeover N1 N2 hence by part (3) of 1.18 we know that N

2 is -prime over N1 + a

soby uniqueness, i.e. 1.17(1), without loss of generality N2 = N2 hence we are done.In general by induction on choose N2, such that N

2,0 is -prime over N

1

A2, N2, is increasing with and N1 N1

N2,. Easily for some , N2, is defined

but not N2,+1. Necessarily N2 is -prime over N1 N

2,. Lastly let a

N2,be such that tp(a, N1 N

2,) does not fork over N1 + a

. Easily N2, is -primeover N1 + a

by (1.17(1)).8) Similar easier proof.9) Let N0 be -prime over A such that B

A

N0, and let N1 be -prime over


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N0 B. By 1.18(1), we know that N1 is -prime over , and by 1.18(10) below N

1

is -prime over A B, hence by 1.17(2) we know that N1, N1 are isomorphic overA B hence without loss of generality N1 = N1 and so N0 = N

0 is as required.

10) By [Sh:c, IV,3.12](3),p.180. 1.18

1.20 Fact. Assume N1 , a : I direct N

2 , a : I (see Definition 1.16)

and A B N1 andI

N2 M.

(1) If = I, then N2

N1

N1 and even N2

N1

I

N1

; and

I implies N2

N1

T

N1

.(2) N2 , a : I is an -decomposition inside M above

BA

iff

N1 , a : I is an -decomposition inside M aboveBA

.

(3) Similarly replacing -decomposition inside M aboveBA

by

-decomposition of M aboveBA

.

Proof. 1) We prove the first statement by induction on g(). If = this isclause (b) by the Definition 1.16 and clause (d) of Definition 1.11(1) (and [Sh:c,V,3.2]). If =, then aN N

1() (by condition (e) of Definition 1.11(1)).

By the induction hypothesis N2()

N1()

N1 and we know N2 is -primary over

N2() N1 ; we know this implies that no p S(N

1 ) orthogonal to N

1 is realized

in N2 henceaN1

N2N1

, so aN1 aN2

henceN1N1

N2N1

hence N1

N1

N2 as required.

The other statements hold by the non-forking calculus (remember if = I

then use tp({N1

: I}, N1

) is orthogonal to N

1

or see details in the proofof 1.21(1)()).2) By Definition 1.16, for = 1, 2 we have: N, a : I is a decomposition

inside C and by assumptionI

N1 N2 M. So for = 1, 2 we have to prove

N, a : I is an -decomposition inside M forBA

assuming this holds for

1 . We have to check Definition 1.11(1).


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22 SAHARON SHELAH

Clauses 1.5(1)(a),(b) for holds because they hold for 1 .Clause 1.5(1)(c) holds as by the assumptions A B N1 N

2, A N

1

and N1 N1N2.

Clauses 1.5(1)(d),(e),(f),(h) holds as N, a : I is a decomposition inside C(for = 1 given, for = 2 easily checked).

Clause 1.5(1)(g) holds as

N1 N2 M is given and M is -saturated.

3) First we do the only if direction; i.e. prove the maximality of N1 , a : I

as an -decomposition inside M forBA

(i.e. condition (i) from 1.11(2)), assuming

it holds for N2 , a : I. If this fails, then for some I\{} anda M, {a : < > I} {a} is independent over N

1 and

a / {a : I} andaN1

N1. Hence, if I for < k then

a = aa : < k realizes over N1 a type orthogonal to N

1, but

N1 N1 , N

1 N

2 and N

1

N1

N2 (see 1.20(1), hence (by [Sh:c, V,2.8])

tp(a, N2 ) N2 hence {a}{a : < k} is independent over N

2 but k,

I for < k were arbitrary so {a} {a : I} is independent over N2

contradicting condition (i) from Definition 1.11(2) for N2 , a : I.

For the other direction use: if the conclusion fails, then for some I\{}anda M\N2\{a : < > I} the set {a : I} {a} is inde-

pendent of N2 and tp(a, N2 ) is orthogonal to N2; let N M be -prime overN2 + a. But N

2 is -prime over N

1 N

2 (by the definition of -direct) so by

NDOP tp(a, N2 ) N1 hence there is a regular q S(N

1 ) such that q tp(a, N

2 ).

Hence some a N realizes q, clearly {a : < > I} {a} is independent

over N2 (and a / {a1 : < > I}) hence over (N

2 , N

1 ) and easily we get

contradiction. 1.20

1.21 Fact. Assume N1 , a1 : I is an -decomposition inside M.

1) IfN1

N2

M, N2

is -prime over andN2

N1

{a1 :< > I} then

()

N2

N1

I

N1

and


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() we can find N2 ( I\{}) such that N2 M, and

N1 , a1 : I

direct N

2 , a

1 : I.

2) If Cb a1

N1 N0 N1 or at least N0 N1 and a

1

N1 N0 whenever

< > I then we can find N0 M and a0 N (for I\{}) such that

N0 , a0 : I

direct N

1 , a

0 : I.

3) In part (2), if in addition we are given B : I such that B is an -finite

subset of N, tp(B, N) does not fork over B

and B

N

0 then we can

demand in the conclusion that I B N0 .

Proof. 1) For proving () let {i : i < i} list the set I such that i j i < j,

so 0 = and without loss of generality for some we have

i {< >:< > I} i [1, ). Now we prove by induction on [1, i)that N2

N1

{N1i : i < }. For = 1 this is assumed. For limit use the local

character of non-forking.If = + 1 [1, ), then by repeated use of [Sh:c, V,3.2] (as {a0 : j [1, )}

is independent over (N1, N2) and N

1 is -saturated and N

1j (j [1, )) is

-prime over N1 + aj ) we know that tp(a , N2

i


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24 SAHARON SHELAH

by 1.18(8).3) Similar. 1.21

1.22 Fact. 1) If N1 , a : I direct N

2 , a : I and both are -

decompositions of M aboveBA

, then

P(N1 , a1 : I, M) = P(N

2 , a

2 : I, M).

Proof. By Defintion 1.11(5) it suffices to prove, for each I\{} that

() for regular p S(M) we have

p N1

& p N1

p N2

& p N2

.

Now consider any regular p S(M): first assume p N1 & p N1 where

I\{} so p N2 (as N1 N

2 and p N

1 ) and we can find a regular

q S(N1 ) such that q p; so as p N1 also q N

1 , now q N

2 (as

N1

N1

N2 and q N1 see [Sh:c, V,2.8]) hence p N

2 .

Second, assume p N2 & pN2 where I\{}; remember N

1, N

1 , N

2 , N

3

are-saturated, N

1 N1

N2 and N2 is -prime over N

1 N

2 and T does not have

DOP. Hence N2 is -minimal over N1 N

2 and every regular q S(N

2 ) is not

orthogonal to N1 or to N2. Also as p N

2 there is a regular q S(N

2 ) not or-

thogonal to p, so as p N2 also q N2; hence by the previous sentence q N

1

hence p N1 . Lastly, as p N2 and N

1 N

2 clearly p N

1 , as required.

1.22

At last we start proving 1.14.

Proof of 1.14. 1) Let N0 C be -primary over A, without loss of generality

N0A

B (but not necessarily N0 M), and let N1 be -primary over N0 B.

Now by 1.18(0) the model N0 is -primary over and by 1.18(1) the model N1 is-primary over hence (by 1.18(10)) is -primary over B, hence without loss ofgenerality N1 M. Let N =: N

0, N = N1, I = {,< 0 >} and a = B.

More exactly a is such that dcl({a}) = dcl(B). Clearly N, a : I is an


. Now apply part (2) of 1.14 proved below.


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2) By 1.13(4) we know N, a : I is an -decomposition inside M, by 1.18(2)we then find J I and N, a for J\I such that N, a : I is an -decomposition of M. By 1.18(3), N, a : J

is an -decomposition of M

above BA where J =: { J : = or 0 J}.3) Part (a) holds by 1.13(2),(3). As for part (b) by 1.13(2) there is N, a : J,an -decomposition of M with I J; easily [0 J I].

1.14(1),(2),(3)

1.23 Fact. If N, a : I

are -decompositions of M aboveBA

, for = 1, 2

and N1 = N2 then P(N

1 , a

1 : I

1, M) = P(N2 , a2 : I

2, M).

Proof. By 1.14(3)(b) we can find J

1

I1

and N

1

, a

1

for J1

\I1

such thatN1 , a1 : J

1 is an -decomposition ofM and moreover we have J1\I1

= & (0 ). Let J2 = I2 (J1\I1) and for J2\I2 let a2 =: a1,

N2 =: N1 . Easily N

2 , a

2 : J

2 is an -decomposition of M. By 1.13(6) we

know that for every regular p S(M) there is (for = 1, 2) a unique (p,) J

such that p N(p,) & p N(p,) (note meaningless). By the uniqueness

of (p,), if (p, 1) J1\I1 then as it can serve as (p, 2) clearly it is (p, 2) so(p, 2) = (p, 1) J1\I1 = J2\I2; similarly (p, 2) J2\I2 (p, 1) J1\I1 and(p, 1) = (p, 2) = . So

() (p, 1) I1\{} (p, 2) I2\{}.

But

() (p,) I\{} p P(N, a : I

, M).

Together we finish. 1.23

We continue proving 1.14.

Proof of 1.14(4). Let A M be -finite, so we can find an -finite B {N : I} such that stp(A, B) stp(A, {N : I}). Hence, there is a finite

non empty I

I such thatd I

, I

is closed under initial segments and B

{N : I

}, so of course stp(A, {N : I}) stp(A, {N : I}).

We can also find B : I such that B is an -finite subset of N, B

=

acl(B) and B {B : I

}, = a B and if I

then B B and tp(B

, N) does not fork over B

. Without loss of generality B B

.

For I\I let B = B where < g() is maximal such that I

, such

exists as g() is finite and I.


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26 SAHARON SHELAH

Let N1 = N and a1 = a for I and without loss of generality J = I hence

J\I = .

Let N2

M be -prime over J\I

N ; letting J\I = {i : i < i

} be

such that [i j i < j] we can find N2,i (for i i

) increasing continuous,

N2,0 = N and N2,i+1 is -prime over N

2,i Ni hence over N

2,i + ai.

Lastly, without loss of generality N2,i = N2.

By 1.18(1),(2) we know N2 is -primary over and (using repeatedly 1.18(6)+ finite character of forking) we have N2

N1

a. By 1.18(4)

(with N1, N2, B

Cb(a/N

1) here standing for N1, N2, A there) we

can find a model N0

such that a N0

N1

and Cb(a

/N1

) B

N0, N0 N

1, N

0 is -primary over and N

1, N

2 are isomorphic over

N0. By 1.21(1) we can for I choose N2 M with N

1 N

2 and N

1 , a

1 :

I direct N2 , a

0 : I. Similarly, by 1.21(2) (here SucI() = {0})

we can choose an -decomposition N0 , a

0 : I with N

0 , a

0 : I

direct

N1 , a1 : I. Moreover, we can demand I

B N0 using 1.21(3).

By 1.13(12) + 1.14(3) we know that N1 , a0 : I is an -decomposition

of M and easily N2 , a0 : I is an -decomposition of M. Now choose

by induction on I an isomorphism f from N1 onto N

2 over N

0 such that

f f and I

f B = idB . For = we have chosen

N0 such that N1 , N

2 are isomorphic over N

0 . For the induction step note that

f() idN0 is an elementary mapping as N2()

N0()

N0 and f() idN0 can

be extended to an isomorphism f from N1 onto N

2 as N

is -primary (in fact

even -minimal) over N() N0 for = 1, 2 (which holds easily). If I

there is no problem to add f B = idB . Now by 1.13(3) the model M

is

-saturated and -primary and -minimal overJ

N =I

N1 ; similarly M is

-primary over I

N2

. Now

f is an elementary mapping from I

N1

ontoI

N2 hence can be extended to an isomorphism f from M into M. Moreover

as stp(A, {B : I

}) stp(A, {N1 : I}), by [Sh:c, CH.XII,4] we have

tp(A, {B : I

} tp(A, {N1 : I} hence tp(A, {B : I

}) hasa unique extension as a complete type over {N1 : I} hence over {N

2 : I}


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so without loss of generality f A = idA . By the -minimality ofM overI

N

(see 1.13(3)), f is onto M, so f is as required. 1.14(4)

We delay the proof of 1.14(5).

Proof of 1.14(6). Let N, a : I

for = 1, 2, be -decompositions of M

aboveBA

, so dcl(a) = dcl(B). Let p S(M), and assume that p P(N

1 , a

1 :

I1, M), i.e. for some I1\{}, (p N) and p N. We shall provethat the situation is similar for = 2; i.e. p P(N2 , a

2 : I

2, M); bysymmetry this suffices.

Let n = g(), choose B : n and d such that:

() A B0,

() B B1,

() a B N1, for n

() B+1B

N1,

()B+1

B+a1(+1) B+1

N1+a1(+1)

,

() d Bn,d

Bn\{d}is regular p, (hence Bn1)

() B is -finite.

[Why such B : n exists? We prove by induction on n that for any I oflength n and -finite B N, there is B : n satisfying () (), () suchthat B Bn. Now there is p

S(N1 ) regular not orthogonal to p, let B1 N1

be an -finite set extending Cb(p). Applying the previous sentence to , B1 we getB : n, let d N realize p

Bn.Now as n > 0, tp(d, Bn) N hence tp(d, Bn) Bn1, hence

tp(d, Bn) tp(N , Bn), hence as tp(d, Bn) is stationary, by [Sh:c, V,1.2](2),p.231,the types tp(d, Bn),tp(N, Bn) are weakly orthogonal so

tp(d, Bn) tp(d, N Bn) henceBn+d

Bn1+a1 Bn+d

N1

+a1.

Now replace Bn by Bn {d} and we finish].

Note that necessarily()+ Bn

Bm

N1m for m n.

[Why? By the non-forking calculus].

()+ BnBm+a1(m+1)

a Bm for m < n.

[Why? As N1m is -saturated].


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28 SAHARON SHELAH

Choose D N2 finite such thatBn

N2+Bdoes not fork over D + B.

[Note: we really mean D N2, not D N1].

We can find N3, -prime over such that A N3 N

2 and D

AN3

and N2 is -prime over N3D

(by 1.18(9)). Hence BnA

N3 and BnB

N3

(by the non-forking calculus). As tp(B, N2) does not fork over A N

3 N

2

by 1.21(2) we can find N3 , a3 (for I

2\{}), such that N3 , a3 : I is an


and N3 , a

3 : I

2 direct N2 , a

3 : I

2and a3 = a

2 (remember dcl(a

2) = dcl(B)). By 1.20(2) we know N

3 , a

3 :

I2 is an -decomposition of M aboveBA

.

By 1.22 it is enough to show p P(N3 , a3 : I

2, M). Let N4 N2 M

be -prime over N3B0. Now by the non-forking calculus BA(N3B0) [why?

because

(a) as said above BnB

N3 but B0 Bn so B0B

N3, and

(b) as BA

N1 and B0 N1 we have B

A

B0 so B0A

B

hence (by (a) + (b) as A B)

(c) B0N3+B

does not fork over A,

also

(d) BA

N3 (as A N3 N

2 and tp(B, N

2) does not fork over A)

putting (c) and (d) together we get

(e)A

{B0, B , N 3}

hence the conclusion].

Hence B N3

B0

so B N3

N4

(by 1.18(6)) and so (as N3

is

-prime over

N3 + dcl(a3) = N

3 + dcl(B)) we have N

4

N3

N3 and by 1.21(1) we can

choose N4 M (for I2\{}), such that N4 , a

3 : I

2 direct N3 , a3 :

I2. So by 1.20(1) N4 , a3 : I

2 is an -decomposition of M aboveBA

hence

a3/N4 does not fork over A but A B0 N

4 so a

3/N

4 does not fork over B0


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and by 1.22 it is enough to prove p P

N4 , a3 : I

2, M

. Now as said above

B

N3

N4 and BA

N3 so together BA

N4, also we have A B0 N4,

hence B B0

N4 and

BnB0+B

BnB0+a3

a B0 (by ()+ above) but a3B0

N4 henceBn

N4+a3

is

-isolated. Also letting Bn = Bn\{d} we have

BnN4+a

3

is -isolated anddBn

B0 (by clause ()), and clearly d

Bn

(N4 Bn) so

dBn

N4. Hence we can find

N5 , a5 : I

5 an -decomposition of M aboveBA

such that N5 = N

4,

dcl(B) = dcl(a5), Bn\{d} N5 and d = a

5 (on d see clause () above)

so d Bn N5.

By 1.23 it is enough to show p P(N5 , a5 : I

5, M) which holds triviallyas tp(d, Bn\{d}) witness. 1.14(6)

Proof of 1.14(5). By 1.8, with A,B,A1, B1 here standing for A1, B1, A2, B2 there,we know that there are B : n, c : 1 < n as there. By 1.18(9) we canchoose N1 such that B0 N

1, N

1

B0

Bn, N1 is -primary over . Then

we choose N1 , a1 : ,< 0 >, < 0, 0 > , . . . , < 0, . . . , 0 >

n, (where

N1< 0, . . . , 0 > n

M and > 0 a1< 0, . . . , 0 >

= c, Bg() N

1 and we choose

N1 by induction on g() being -prime over N1 a

1 hence a

1/N

1 does not

fork over Bg() hence N1 is -prime also over N

1 + B

g(). So N

1 , a

1 :

{< > ,.. . } is an -decomposition inside M forB1A1

. Now apply first 1.14(2) and

then 1.14(6).

Proof of 1.14(7). Should be easy. Note that

()1 for noB

A

do we have

BA

bB

A

Why? By the definition of Depth zero.

()2 ifBA


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30 SAHARON SHELAH

() for noB1A1

,B2A2

do we have

BA a B1

A1


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Proof. 1) (c) (b).

By 1.14(1) there is an -decomposition of M1 aboveBA

. By clause (c) it is

also an -decomposition of M2 above BA, just as needed for clause (b).

(b) (a)Let N, a : I be as said in clause (b). By 1.14(3)(b) we can find J1, I J1

and N, a (for J1\I) such that N, a : J1 is an -decomposition ofM1and J1\I (0) > 0. Then we can find J2, J1 J2 and N, a (for J2\J1)such that N : J2 is an -decomposition of M2 (by 1.14(2)). By 1.14(3)(b), J2\I (0) > 0. So I\{} SucJ2() = SucI(), hence

() if I\{} and q S(N) is regular orthogonal to N then the station-arization of q in S(M1) is not realized in M2.

Now if p PBA, M1 then p S(M1) is regular and (see 1.14(1), 1.11(5)) forsome I\{}, p N , p N, so there is a regular q S(N) not orthogonal top. Now no c M2 realizes the stationarization of q over M1 (by () above), hencethis applies to p, too.

(a) (c)Let N, a : I be an -decomposition of M1 above

BA

. We can find

N, a : J an -decomposition of M1 such that I J and J\I (0) > 0 (by 1.14(3)(b)), so M is -prime over {N : J}. We should

check that N : a : I it is also an -decomposition of M2 above

BA

, i.e.

Definition 1.11(1),(2). Now in 1.11(1), clauses (a)-(h) are immediate, so let uscheck clause (i) (in 1.11(2)). Let I\{}, now is {a : I} reallymaximal (among independent over N sets of elements of M2 realizing a type fromP = {p S(N) : p orthogonal to N})?. This should be clear from clause (a)(and basic properties of dependencies and regular types).2) By 1.14(5).3) Left to the reader. 1.24

1.25 Conclusion: Assume M1 M2 are -saturated andB1A1

B2A2

both in

(M1). If clause (a) (equivalently (b) or (c)) of 1.24 holds for B1A1, M1, M2 thenthey hold for B2A2, M1, M2.

Proof. By 1.24(2), clause (a) forB1A1

, M1, M2 implies clause (a) for

B2A2

, M1, M2.

1.25


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32 SAHARON SHELAH

1.26 Claim. IfB1A1

(M) and N, a : I is an -decomposition of M

aboveB1A1

and M M is -saturated and

I

N M and is an ordinal

then

tpB1

A1

, M

= tpB1

A1

, M

Proof. We prove this by induction on (for all B,A, N, a : I, I , M and M

as above). We can find an -decomposition N, a : J of M with I J (by1.13(4) + 1.13(2)) such that J\I = and 0 and so M is -prime

over JN and also over M {N : J\I}.

Case 0: = 0.Trivial.

Case 1: is a limit ordinal.Trivial by induction hypothesis (and the definition of tp).

Case 2: = + 1.

We can find M M which is -prime over IN, so as equality is transitive

it is enough to prove

tp

B1A1

, M

= tp

B1A1

, M

and

tp

B1A1

, M

= tp

B1A1

, M

.

By symmetry, this means that it is enough to prove the statement when M is-prime over

I

N.

Looking at the definition of tp+1 and remembering the induction hypothesisour problems are as follows:First component of tp:


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givenB1A1

aB2A2

, B2 M, it suffices to find

B3A3

such that:

() there is f AUT(C) such that: f B1 = idB1 , f(A2) = A3,

f(B2) = B3 and B3 M

and tp [B2A2, M] = tp [B3A3, M](pedantically we should replace B, A by indexed sets).We can find J, M such that:

(i) I J J, |J\I| < 0, J closed under initial segments,(ii) M M is -prime over M

{N : J\I}

(iii) B2 M.

The induction hypothesis for applies, and gives

tpB2A2, M = tpB2A2, M.

By 1.14(4) there is g, an isomorphism from M onto M such that g B1 = id.So clearly g(B2) M

hence

tpB2

A2

, M

= tpg(B2)

g(A2)

, M

.

SoB3A3

=: gA2B2

is as required.

Second component for tp:

So we are given , a tp type, (and we assign the lower part as B) and we have toprove that the dimension in M and in M are the same, i.e.dim(I, M) = dim(I, M), where: I = {c M : = tp

cB1

, M

} and

I =

c M : = tp

cB1

, M

.

Let p be such that: tp

cB1

, M

= p = cB1 . Necessarily p A1 and p is

regular (and stationary).Clearly I I, so without loss of generality I = hence p is really well defined,

now

() for every c I for some k < , c M realizing p for < k we have c

depends on {c0, c1, . . . , c

k1} over B1.

[Why? Clearly p N (as B1A1

N and p A1) hence

tp(

J\I

N, N) p hence


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34 SAHARON SHELAH

tp(

J\I

N, M) p, but M is -prime over M

J\I

N hence

by [Sh:c, V,3.2,p.250] for no c M\M is tp(c, M) a stationarization of

p hence by [Sh:c, V,1.16](3) clearly () follows].

If the type p has depth zero, then (by 1.14(7)):

I = {c M : tp(c, B) = p} and

I = {c M : tp(c, B) = p}.

Now we have to prove dim(I, A) = dim(I, A), as A is -finite and M, M are-saturated and I I clearly

0 dim(I, A) dim(I, A). Now the equality

follows by () above.So we can assume p has depth > zero, hence (by [Sh:c, X,7.2]) that the type

p is trivial; hence, see [Sh:c, X7.3], in () without loss of generality k = 1 anddependency is an equivalence relation, so for same dimension it suffices to provethat every equivalence class (in M i.e. in I) is representable in M i.e. in I. Bythe remark on () in the previous sentence (d1 I)(d2 I)

d1B1

d2

. So it

is enough to prove that:

if d1, d2 M realize same type over B1, which is (stationary and) regular,

and are dependent over B1 and d1 M

then there is d

2 M

such thatd2B1+d1

= d2B1+d1 and tp

B1+d2B1

, M

= tpB1+d

2

B1

, M

.

Let M0 = N. There are J, M1, M

+1 such that

()1(i) J J is finite (and of course closed under initial segments)

(ii) J, 0 / J

(iii) M1 M is -prime over {N : J}

(iv) M+1 M is -prime over M1 M (and M1

M0

M)

(v) d2 M+1 .

Now the tripleB1+d2B1

, M1, M satisfies the demand on

B1A1

, M, M (because

B1A1

B1+d2B1

,by 1.25. Hence by the induction hypothesis we know that

tpB1 + d2

B1

, M

= tpB1 + d2

B1

, M+1

.


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By 1.29(4) there is an isomorphism f from M+1 onto M which is the identity on

B1 + d1; let d2 = f(d2) so:

tpB1 + d2B1 , M+1 = tpB1 + d

2B1

, M.Together

tpB1 + d2

B1

, M

= tpB1 + d2

B1

, M

.

As {d1, d2} is not independent over B1, also {f(d1), f(d2)} = {d1, f(d2)} is notindependent over B1, hence, as p is regular

() {d2, f(d2)} is not independent over B2.

Together we have proved

, hence finished proving the equality of the secondcomponent.

Third component: Trivial.So we have finished the induction step, hence the proof. 1.26

1.27 Claim. 1) Suppose M is -saturated, A B M, BA ,

2

=1[A A

M],A = ac(A), A are -finite,

A1A =

A2A , B

A

A1 and BA

A2.

Then tpA1BA1

, M

= tpA2BA2

, M

for any ordinal .

2) Suppose M is -saturated, B M,BA

,

2=1

[A A M], A = ac(A),

B = ac(B), A = ac(A), A is -finite,A1A =

A2A , B

A

A1, BA

A2, f : A1onto A2

an elementary mapping, f A = idA, g f idB , g elementary mapping from

B1 = ac(B A1) onto B2 = ac(B A2).

Then g

tpB1A1

, M

= tpB2A2

, M

for any ordinal .

3) Assume that

(a) A = acl(A) B = acl(B) M for = 1, 2

(b) A A+ acl(A

+ ) M

for = 1, 2


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36 SAHARON SHELAH

(c) BA

A+ for = 1, 2

(d) f is an elementary mapping from A1 onto A2

(e) g is an elementary mapping from A+1 onto A+2(f) f A1 = g A1

(g) h is an elementary mapping from B+1 = acl(B1 A+1 ) onto B

+2 = acl(B2

A+2 ) extending f and g

(h) f(tp[B1A1

, M1]) = tp[

B2A2

, M2].

Then h(tp[B+1A+1

, M1]) = tp[

B+2A+2

, M2].

Proof. 1) Follows from part (2).2) We can find A3 M such that:

(i) A3A =A1A

(ii) A3A

(B A1 A2).

Hence without loss of generality A1B

A2 and evenA

{B, A1, A2}. Now we can find

N, an -prime model over , N M, A N and (B A1 A2)A

N

(e.g. choose {A1 A

i B

: } M indiscernible over A, A1 = A1, A

2 =

A2, B = B and let N M be -primary over

n } and J = I {< 1 >, < 2 >}. LetN2 M

be -prime over NN. By 1.21 there is N2 , a : I an -

decomposition of M aboveBA

such that N, a : I direct N2 , a : I.

Let M M be -prime overI

N2 and M M be -prime over

I

N. So

M

M

M and M

is -prime over M

N N.Now by 1.26 we have tp

BA

, M

= tp

BA

, M

for = 1, 2 hence it suffices

to find an automorphism ofM extending g. Let B+ = ac(NB), A = ac(B A), let a list A

be such that a2 = g(a1). Clearly tp(a, B

+) does not fork overA B and ac(B) = B and so stp(a1, B+) = stp(a2, B+). Also tp(A2, B+ A1)does not fork over A hence tp(a2, B

+ a1) does not fork over A B+ hence

{a1, a2} is independent over B+ hence there is an elementary mapping g+ from


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ac(B+ a1) onto ac(B+ a2), g

+ idB+ g and even g = g+ (g+)1 is an

elementary embedding.

Let a1 lists ac(N A1) so clearly a2 =: g

+(a1) list ac(N A2). Clearly

g

(a1 a

2) is an elementary mapping from a

1 a

2 onto itself. Now N

2 is -

primary over NA1A2 and NA1A2 a1a2 ac(NA1A2) so by

1.18(10) N2 is -primary over N a1 a

2 hence we can extend g

(a1 a2)

to an automorphism h of N2 so clearly h N = idN. Let a

+1 list

ac(B+ A1) and a+2 = g

+(a+1 ). So tp(a+ , N

2) does not fork over a

1( N

2)

and ac(a1) = Rang(a1) (= ac(NA1)) and h a

1 = g

+ a1 hence hg+

is an elementary embedding. Remember g+ is the identity on B+ = ac(N B),and tp(N, N

2) does not fork over N hence tp(N, B

+ N2) does notfork over B+, so as ac(B+) = B+ necessarily (hg

+) idN is an elementaryembedding. But this mapping has domain and range including N N

2 and

included in N2

, but the latter is -primary and -minimal over the former.Hence (h g+) idN can be extended to an automorphism of N

2 which

we call h.

Now we define by induction on n [2, ) for every I of length n, anautomorphism h ofN

2 extending h idN , which exists as N

2 is -primary over

N2N (and N2

N

N). NowI

h is an elementary mapping (as N2 : I

is a non-forking tree; i.e. 1.13(10)), with domain and rangeI

N2 hence can be

extended to an automorphism h of M, (we can demand h M = idM but

not necessarily). So as h

extends g, the conclusion follows.3) Similarly to (2). 1.27

1.28 Claim. 1) For every = t pBA

, M

, and a, b listing A, B respectivelythere is = (xA, xB) L,(q.d.) of depth such that:

tpB

A

, M

= M |= [a, b].

2) Assume M1,M2 of 1.4 holds as exemplified by the familyF and B

A (M1)and g F, Dom(g) = B; and an ordinal thentp

B

A

, M

= tp

g(B)

g(A)

, M2

.

3) Similarly for tp([A], M), tp[M].


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38 SAHARON SHELAH

Proof. Straightforward (remember we assume that every first order formula isequivalent to a predicate). 1.28

1.29 Proof of Theorem 1.2. [The proof does not require that the M are -saturated, but only that 1.27, 1.28 hold except in constructing g() (see 14, 15in 1.30(E), we could instead use NOTOP].

So suppose

()0 M1 L,(d.q.) M

2 or (at least)

M1,M2 from 1.4 holds.

We shall prove M1 = M2. By 1.28 (i.e. by 1.28(1) if the first possibility in ()0holds and by 1.28(2) if the second possibility in ()0 holds)

()1 tp[M1] = tp[M2].

So it suffices to prove:

1.30 Claim. Assume that T is countable. If M1, M2 are -saturated models (ofT, T as in 1.5), then:

()1 M1 = M2.

Proof. Let Wk, W

k : k < be a partition of to infinite sets (so pairwise disjoint).1.31 Explanation: (If seems opaque, the reader may return to it after reading partsof the proof).

We shall now define an approximation to a decomposition. That is we are ap-proximating a non-forking tree N, a

: I

of countable elementary submodels

of M for = 1, 2 and f : I such that f an isomorphism from N

1 onto N

2

increasing with such that M is -prime overI

N.

In the approximation Y we have:

() I approximating I,[it will not be InOrd but we may discover more immediate successorsto each I; as the approximation to N improves we have more regulartypes, but some member of I will be later will drop]

() A approximates N and is -finite

() a is the a (if survives, i.e. will not be dropped)


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() B, b,m expresses commitments on constructing A

: we promise

B N and B

is countable; b

,m for m < list B

(so in the choice

B M there is some arbitrariness)

() f approximate f

() p,m also expresses commitments on the construction.

Since there are infinitely many commitments that we must meet in a constructionof length and we would like many chances to meet each of them, the sets Wk, W

k

are introduced as a further bookkeeping device. At stage n in the construction wewill deal e.g. with the b,m for that are appropriate and for m Wk for some

k < n and analogously for p,m and the Wk.

Note that while the A satisfy the independence properties of a decomposition,

the B do not and may well intersect non-trivially. Nevertheless, a conflict arisesif an a falls into B

since the a

are supposed to represent independent

elements realizing regular types over the model approximated by A but now a

is in that model. This problem is addressed by pruning < i > from the tree I.

1.32 Definition. An approximation Y to an isomorphism consist of:

(a) natural numbers n, k and index set: I nOrd(and n minimal)

(b) A, B, a, b,m : I and m k


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40 SAHARON SHELAH

(5) if I\{}, thena

A()

is a (stationary) regular type and a A; if in

addition g() > 1 thena

A()

A() (note that we may decide a be

not defined or A)

(6)A

A

+aa A if I,g() > 0

(7) if I, not -maximal in I, then the set {a : I and = } is a

maximal family of elements realizing over A regular types A() (when

is defined), independent over

A, B

, (and we can add: if

1 = 2 = and

a1A

a2A

then a1/A = a2/A)

(8) f is an elementary map from A1 onto A

2

(9) f() f when I,g() > 0(10) f(a

1) = a

2

(11) () f

tp

A1A1()

, M1

= tp

A2A2()

, M2

when I\{}

() f

tp

A1, M1

= tp

A2, M2

(12) B M moreover, B na M

, i.e., if a N, b M\B and M

|=

(b, a) there for some b B, |= (b, a) and b / acl(a) b / acl(A)

(13) p,m : m k


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Proof. Let I = {}, A = ac(), k = 1 and then choose countable B to

satisfy condition (12) and then choose f, pk, b

,m (for k W

0 and m W0) as

required.

1.35 Main Fact. For any approximation Y, i

k 0 b

A() [Y]

a A() [Y]

(b) one of the conditions (i),(ii) listed below holds for b

(i) b = b(),i [Y] and

g() > 0 bA

() [Y]

a A() [Y] or

(ii) for no b is (i) satisfied (so g() > 0) and b M(),

b,i

A() [Y]

b and g() > 0 bA() [Y]

a A()

[Y]

4recall that i is part of the information given in the main fact, and of course, k is uniquelydetermined by i


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42 SAHARON SHELAH

()2 assume IY, g() = m, k < kY and i W

k then we have:

(a) if p

()

,i is realized by some b M()

such that

Rk

b

A() [Y]

, L,

= Rp(),i , L,

and

g() > 0 b

A() [Y]

a A() [Y]

then for some such b we have

A() [Z] = acA

() [Y] {b}

(b) if the assumption of clause (a) fails but p(),i is realized

by some b M()\A() such that

g() > 0 b

A() [Y]

a A() [Y]

then for some such b we have

A() [Z] = ac

A() [Y] {b}

() If IY and g() = m, then B[Z] = {b,j [Y] : j {Wk : k < k

Z}}

is a countable subset ofM, containing {B [Z] : and Y} B[Y],

with B[Z] M moreover B[Z] na M

i.e. if a B[Z], (x, y) is

first order and (x M\ac(a)) (x, a) then (x B[Z]\ac(a))(x, a))

and {a[Y] : IY and a[Y] / B

[Z]} is independent over

(B[Z], A[Y])

() if IY, g() > m, then IZ a(m+1)

[Y] / Bm

[Z]

() if IY IZ , g() > m then A[Z] = ac(A[Y] A

m[Z]) and

B[Z] = B[Y]

() if IZ\IY then IY and g() = m + 1

() {p,i[Z] : i WkZ1

} is rich enough, e.g. include all finite types over A

() {b,i : i Wkz1} list B[Z], each appearing infinitely often.


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Proof. First we choose A() [Z] for I of length m according to condition

() = ()1 + ()2. (Note: one of the clauses ()1, ()2 necessarily holds trivially as

kWk k

Wk = ).

Second, we choose (for such ) an elementary mapping fZ extending fY and a set

A3() [Z] M3() satisfying fZ is from A

1[Z] onto A

3() [Z] such that

()2 if m > 0, then

fZ

tp A1 [Z]

A1

[Y]

, M1

= tp A2[Z]

A2

[Y]

, M2

()3 if m = 0, then fZ

tp

A1[Z], M1

= tp

A2[Z], M2

.

[Why possible? If we ask just the equality of tp

for an ordinal , this follows by

the first component of tp+1. But (overshooting) for

(M1 + M2)|T|+

,equality of tp implies equality of tp].

Third, we choose B[Z] for IY, g() = m according to condition () (here weuse the countability of the language, you can do it by extending it times) in bothsides, i.e. for = 1, 2.

Fourth, let I = { I : if g() > m then a(m+1)[Y] / Bm[Z]} (this will be

IY IZ).

Fifth, we choose A[Z] for I: if g() < m, let A[Z] = A

[Y], if g() = m

this was done, lastly if g() > m, let A[Z] = ac A[Y] Am[Z].Sixth, by induction on k nY we choose f

Z for I

of length k: if g() < m,

let fZ = fY , if g() = m this was done, lastly if g() > m choose an elementary

mapping from A1 onto A2 extending f

Y f

Z (possible as f

Y f

2 is an elementary

mapping and Dom(fY ) Dom(fZ) = A

() , Dom(f

Y )

A()

Dom(fZ)

and A()

= ac(A()

).) Now fZ satisfies clause (11) of Definition 1.32 when

g() > m by applying 1.27(3).

Seventh, for I, of length m < nZ , let v =: { : I}, and we choose{a1[Z] : u}, [ u / I], a set of elements of M

1 realizing

(stationary) regular types over A1[Z], orthogonal to A[Y] when g() > 0, such

that it is independent over

{a1[Y] : I} B1[Z], A

1[Z]

and max-

imal under those restrictions. Without loss of generality sup(v) < min(u) andfor 1 v u and 2 u we have:


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44 SAHARON SHELAH

()1 if (for the given 2 and ) 1 is minimal such thata1

[Z]

A1[Z]

a1[Z]

A1[Z]then

a1[Z]

A1[Z]=

a1[Z]

A1[Z]

()2 if 1 < 21 and a11[Z]/A1[Z] = a12[Z]/A1[Z] and for some b M1

realizinga1[Z]

A1[Z]we have

b

A1[Z]

a1 and

tp

b

A1

, M

= tp

a1A1

, M

and 1 is minimal (for the given 2 and ) then

tp

a1A1

, M

= tp

a1A1

, M

.

Easily (as in [Sh:c, X]) if u and I then

a1[Z]

A1[Z]

a1[Y]

A1[Y].

For u let A1[Z] = ac

A1[Y] {a

1[Z]}

.

Eighth, by the second component in the definition of tp+1 (see Definition 1.9) wecan choose (for u) a

2[Z], A

2[Z] and then f

Z as required (see (7)

of Definition 1.32).

Ninth and lastly, we let IZ = I { < >: I, g() = m < nZ and u}

and we choose B for IZ\IY and the p,i, b

,j as required (also in other case

left).1.35

1.36 Finishing the Proof of 1.11. We define by induction on n < an approximationYn = Y(n). Let Y0 be the trivial one (as in observation 1.30(C)).

Yn+1 is gotten from Yn as in 1.35 for mn, in n, n() {1, 2} defined by

reasonable bookkeeping (so in

k


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B[] =nOrd is closed under initial segments1 for I

, B[Yn] : n < and I[Yn] is an increasing sequence of

na-elementary submodels of M

[Why? By clause (12) of Definition 1.32, Main Fact 1.35, clauses ( )(a), (), ().]

hence2 for I

, B[] na M.

Also3 I

B [] B[].

[Why? Because for infinitely many n, mn = g() and clause () of Main Fact 1.35].4 if I[Yn1 ] I

, = and n1 n2 then

A[Yn1 ]

A[Yn1 ]

A [Yn2 ].

[Why? Prove by induction on n2 (using the non-forking calculus), for n2 = n1 this

is trivial, so assume n2 > n1. Ifm(n21) > g() we have A [Yn2 ] = A [Yn21] (see1.35, clause ()(a) and we have nothing to prove. If m(n21) < g() then we note

that A [Yn2 ] = acl(A[Yn21]A

m(n21)

[Yn2 ]) and A [Yn21]

Am(n21)

Am(n21)[Yn2 ]

(as I[Yn2 ], by 1.35 clause () last phrase) and now use clauses (5), (6) of Defi-nition 1.35. Lastly if m(n21) = g() again use I[Yn2 ] by 1.35, clause (), lastphrase].

5 if I[Yn1 ] I, = and n1 n2 then

A

[]A[] + a

[]

a A.

[Why? By clause (6) of Definition 1.32, and orthogonality calculus].6 if I

, then A[] B[] M

moreover7 A

()[] na B

[] na M

.


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46 SAHARON SHELAH

[Why? The second relation holds by

2. The first relation we prove by inductionon g(); clearly A[] = ac(A

[]) because A

[Yn] increases with n by 1.35 and

A[Yn] = acl(A[Yn]) by clause (3) of Definition 1.32. We prove A

()[] na

B[] by induction on m = g(), so suppose this is true for every m < m, m =g(), I, let (x) be a formula with parameters in A[] realized in M

as

above say by b M. As A[Yn] : n < , Yn is increasing with union A[],

clearly for some n we have b

A[Yn]

A[].

So {(x)} = p,i for some i and for some n > n defining Yn+1 we have used

1.35 with ((), i , m) there being (,i,g()) here, hence we consider clause ()2 of1.35. So the case left is when the assumption of both clauses (a) and (b) of ()2

fail, so we have g() > 0 and

b / A[Yn ], b M |= [b] b

A[Yn ] A[Yn ].

We can now use the induction hypothesis (and [BeSh 307, 5.3,p.292]).]

8 if I

and = 1, 2, then

{a[] : I} is a maximal subset of

{c M :c

A[]regular, c

A[]

B[] and g() > 0 c

A[] A []}

independent over (A[], B[]).

[Why? Note clause (7) of Definition 1.32 and clause () of Main Fact 1.35].9 A

[] = B

[]

[Why? By the bookkeeping every b B[] is considered for addition to A[]

see 1.35, clause ()1, subclause (b)(i) and for there is nothing to stop us].

10 if I\ and p S(A[]) is regular orthogonal to A[] then

B[]

A[] p.

[Why? If not, as A[] na B[] by [BeSh 307, Th.B,p.277] there is

c B[]\A[] such that:

cA[]

is p. As c B[] =n


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the triple (in, n, mn) is such that n = , mn = g() and b,in

= c, so by clause

()1(b)(ii) of 1.35 we have c A[Yn] A[].]

11 if I, {1, 2} then {a : I} is a maximal subset of{c M : cA [] regular, A [] when meaningful} independent overA[].

[Why? If not, then for some c M, {a : I} {c} is indepen-

dent over A[] and tp(c, A[]) is regular (and stationary). Hence by 10 we

have {a[Yn] : I} {c} is independent over (A[], B

[]). Now for

large enough n we have c

A[Yn]

A[] and by 10 we have c

A[Yn]

B[], hence

c A[]B[Yn], and so by Definition 1.32(7) {c} {a

[Yn] : I[Yn]} is

not independent over (A[Yn], B[Yn]), but {a

[Yn] : I[Yn]} is inde-

pendent over (A[Yn], B[Yn]). So there is a finite set w of ordinals such that

w I[Yn] and {c} {a[Yn] : w} is not independent over

(A[Yn], B[Yn]), and without loss of generality w is minimal. Let n1 [n, ) be

such that w & a B[] a

B

[Yn1 ]; clearly exist as w is finite

and let u = { w : a / B[]; clearly u < > I

. Now

{a[] : I} B[] includes {a

[Yn] : w}, easy contradiction to

the second sentence above.]

12 f

=

m


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48 SAHARON SHELAH

(c) () is maximal, i.e., we cannot find d1() such that the demand in

(a) holds for () + 1.

[Why? We can try to choose by induction on , a member d1 of M1\

I[]

{d1 :

< } such that tp(d1,

I[]

A[] {d1 : < }) is -isolated and F

0

-isolated.

So for some (), d1 is well defined iff < () (as < d1 = d

1 M

1).

Now choose by induction on < (), d2 M2 as required above, possible by M2i

being -saturated (see [Sh:c, XII,2.1,p.591], [Sh:c, IV,3.10,p.179].]

15 Dom(g()), Rang(g()) are universes of elementary submodels of M1, M2

respectively called M1, M2 respectively.

[Why? See [Sh:c, XII,1.2](2),p.591 and the proof of 14.Alternatively, choose a formula (x, a) such that:

(a) a Dom(g()) and |= x(x, a) but no b Dom(g()) satisfy (x, a)

(b) under clause (a), Rk((x, a),L|T|, ) is minimal(or just has no extension in S(Dom(g())) forking over a).

Let {(x, y) : < } list that L(T)-formulas and we choose by induction on as

formula n(x, an) such that:

(i) a Dom(g())

(ii) |= (x)n(x, an)

(iii) n+1(x, an+1) n(x, an)

(iv) 0(x, a0) = (x, a)

(v) for any formula (x, a) satisfying the demands on n+1(x, an+1) we haveRk(n+1(x, an+1), {n(x, yn)}, 2) < Rk(

(x, a), {n(x, y)}, 2)(on this rank see [Sh:c, II,2]).

So p = {n(x, an) : n < } has an extension in S(Dom(g())) call it q. Now qis -isolated because (x, a) q S(Dom(g()). For every n, n+1(x, an) q {n(x, yn)} by clause (v) above so as n+1(x, an) q and this holds for every nclearly q is F0-isolated.

16 If M = M then for some d M\M

,

dM

is regular.


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[Why? By [BeSh 307, Th.5.9,p.298] as N na M by

7].

17 if M

= M then for some I, there is d M\M such that

dA[]

is

regular, d A[]

M and g() > 0 dA[] A [].[Why? By [Sh:c, XII,1.4,p.529] every non-algebraic p S(M) is not orthogonalto some A[] so by 16 we can choose I

and d M\M such thatdM

saharon shelah- characterizing an ℵ-epsilon saturated model of superstable ndop theories by its...

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